One mole of a substance is defined by Avogadro as consisting of 6.022 x 1023 atoms. This is Avogadro's number. To calculate the number of atoms in two moles of sodium, use dimensional analysis. 2.0 moles Na x 6.022⋅1023g1mol=1.20⋅1024 atoms of Na
Answer: Use Avogadro's number, NA = 6.02 × 1023 atoms/mol.
Explanation:
Answer:
beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.
Explanation:
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
